Question

Find
${\int }_0^{1/4\pi }ln(1+\tan x)dx$.

Answer 1

$I={\int }_0^{\frac{\pi }{4}}\ln (1+\tan x)dx$

Use king

$I={\int }_0^{\frac{\pi }{4}}\ln (1+\tan (\frac{\pi }{4}-4))dx$

$={\int }_0^{\frac{\pi }{4}}\ln (1+\frac{1-\tan x}{1+\tan x})dx$

$={\int }_0^{\frac{\pi }{4}}\ln \left(\frac{1+\tan x+1-\tan x}{1+\tan x}\right)dx$

$={\int }_0^{\frac{\pi }{4}}\ln \left(\frac{2}{1+\tan x}\right)dx$

$2I={\int }_0^{\frac{\pi }{4}}\ln \left(\frac{2}{1+\tan x}\right)+\ln (1+\tan x)dx$

$2I={\int }_0^{\frac{\pi }{4}}\ln \left(2\right)dx$

$2I=\ln \left(2\right)(\frac{\pi }{4}-0)$

$I=\frac{\pi }{8}\ln \left(2\right)$